Are you interested about values [itex]a,b,c\in\mathbb{N}[/itex] or [itex]\in\mathbb{R}[/itex]?

With natural numbers, there is an equivalent problem with small cubes of fixed size. Suppose you have c^3 small cubes, piled into a one bigger cube. Can you take these c^3 small cubes, and form two new big cubes that contain precisely all these small cubes? The Fermat's last theorem says that you cannot.

With real numbers, an equation x^3+y^3=1 describes a surface of a L^p ball with value p=3. Frankly, I still have not understood why real analysis deals so much with L^p spaces. I don't know what's their significance, yet

Ok tell me if theres a problem with this but I think this is the only way to solve it.
cube root 5^3+cube root 5^3= cube root 10^3
the cube root just cancels out the cubing so the a and b variables need to be the same and the c variable needs to be double the a or b.
I don't see what use this would have but thats the only way I've found to solve it.

There is no solution. No 2 (natural number) cubes add up to another cube. In fact no solutions exist in x^n + y^n =z^n, where n is any number greater then 2, as stated by Fermat's last theorem (as proved by Andrew Wiles).

bananana you can't cancel out like this. One way to show this is wrong is

I'm still not seeing why a cube root number wouldn't cancel out the cube. I know it has no practical application but that equation i gave would work. what you left out in your proof is the cube root so it should be the (cube root of 5) to the third power. This also got me thinking about the ratio of three dementional object sides to each other, does anyone know some equations along with their application in regards to shape.

I'm still not seeing why a cube root number wouldn't cancel out the cube. I know it has no practical application but that equation i gave would work. what you left out in your proof is the cube root so it should be the (cube root of 5) to the third power. This also got me thinking about the ratio of three dementional object sides to each other, does anyone know some equations along with their application in regards to shape.

ok so I actually did this on a calculator, feel free to do so yourself, and the equation works. Now the problem i see with it is that it takes away from the "spirit" of the equation in the sense that it is just canceling out the main part of the equation and therefore basically destroying it. It reminds me of those verbal equations kids would say like, give me your favorite number then multiply by 2 add this and that, and the number would end up being the original through inverse operations. Basically i proved that there is a way to solve it but its pointless.

My point is yes, for yours it works because basically what you are doing is picking numbers that will work for your method, but you can not apply this method to everything because it will not work for all cases in which a +b=c would also work for the cube of each.

Also the reason the cube root is irrelevantly is because every number has a cube root, so if we have the number a, the cube root could be x ( cube root a=x) and then simplify the problem so that x^3 + y^3=z^3, and I I showed this does not work. Look up fermats last theorem it disproves it all.